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Sign problem with potential

  1. Feb 16, 2014 #1
    I've been having a problem with the sign of potential. Electric potential as I know it is defined as V = -[itex]\int_{C}\vec{E}\bullet\vec{dl}[/itex] where C is the path from a location defined as zero potential to the location you are measuring the potential at. Now I want to run through this really quick with a simple problem such as the potential of a positive point charge. [itex]\vec{E}[/itex]=[itex]\frac{q\hat{r}}{4\pi\epsilon_{o}\tilde{r}^{2}}[/itex] for a point charge (tilde as an integration variable) and because we are moving in a path from infinity (the point defined as zero potential in this case) to some point r, [itex]\vec{dl}[/itex] should equal -d[itex]\tilde{r}\hat{r}[/itex]. Hence V = [itex]\int^{r}_{∞}E\left(\tilde{r}\right)d\tilde{r}[/itex] = -[itex]\frac{q\hat{r}}{4\pi\epsilon_{o}r}[/itex] which is the negative of what it should be.

    What's wrong here? I've wondered for months and gotten unsatisfactory explanations from TAs and professors.

    Thanks!
     
  2. jcsd
  3. Feb 16, 2014 #2
    What is going on is essentially this: Suppose you are integrating a constant (equal to unity for convenience) from x=1 to x=0 along the x axis. Would you say that because we are moving in a negative direction, [itex]dl=-dx[/itex], and thus
    [tex]\int_1^0 dl = -\int_1^0 dx = 1[/tex]
    or would you just say that the integral is over a change in x with defined limits
    [tex]\int_1^0 dx = -1[/tex]
     
  4. Feb 17, 2014 #3
    I kinda see what you're getting at, but still struggling a bit. Intuitively, I see the path from infinity to r as being a negative path (moving in a path against a field), but am I simply adding a negative that is already supplied by the integral?

    What I got from what you said is this,

    [itex]d\vec{l}[/itex] = [itex]dr\hat{r}+d\theta\hat{\theta}+d\phi\hat{\phi}[/itex]
    [itex]\vec{E}\bullet d\vec{l}=Edr[/itex]

    And the negative sign from the path being intuitively negative is provided by the integral?

    Another thread of thought,

    [itex]dl=-dr[/itex]

    [itex]\vec{E}\bullet d\vec{l}=-Edl=Edr[/itex] (assuming E and dl antiparallel from my previous example)

    vs

    [itex]d\vec{l}=-dr\hat{r}[/itex]

    [itex]\vec{E}\bullet d\vec{l}=-Edr[/itex]

    Or wait!

    [itex]d\vec{l}=dr\hat{r}\neq-dr\hat{r}[/itex]

    because [itex]d\vec{l}[/itex] points in the [itex]-\hat{r}[/itex] direction and dl=-dr! It's a double negative.

    Thanks for the help, I think I got it. I'm gonna leave all my work up because I am still interested if

    [itex]d\vec{l}[/itex] = [itex]dr\hat{r}+d\theta\hat{\theta}+d\phi\hat{\phi}[/itex]
    [itex]\vec{E}\bullet d\vec{l}=Edr[/itex]

    is a proper way to think about the problem.
     
  5. Feb 17, 2014 #4
    In spherical polar coordinates

    [itex]\vec{dl} = dr\hat{r}+rd\theta\hat{\theta}+rsin\theta d\phi\hat{\phi}[/itex]

    [itex]\int\vec{E}\bullet d\vec{l}=\int Edr[/itex]

    dr is the change in the radial displacement of the test charge.It is positive if test charge is moved away from the origin(point charge).OTOH it is negative if test charge moves towards the origin(point charge).
     
    Last edited: Feb 17, 2014
  6. Feb 17, 2014 #5
    Whoops. Got spherical polars wrong. That's embarrassing.

    Thanks for the help!
     
  7. Feb 17, 2014 #6

    Philip Wood

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    Gold Member

    In a nutshell, you should have retained the minus sign arising from [itex]dV = - E_r dr[/itex]. This then cancels with the minus from the integration of [itex]\frac{1}{r^2} dr[/itex].

    We must be kindred spirits, because I also spent a lot of time trying to justify to myself what I was doing when I first taught this bit of work.

    If all you want is a clear derivation, and you don't care if it's not re-inforcing formal techniques, then simply calculate the work done by the field on a test charge, per unit charge, as it goes from r to infinity. That gives you the potential at r, as the work the field can do on a charge, per unit charge, as it goes from r to infinity is by definition the potential at r.
     
  8. Feb 17, 2014 #7
    Hi Philip...

    You are quite right . Instead of work done by external agent in bringing test charge from infinity to 'r' , it should be taught as work done by electric field in moving test charge from 'r' to infinity.This avoids the confusion arising from the sign of displacement .

    Again ,learning the definition of potential in terms of work done by electric field is comprehensible.
     
  9. Feb 17, 2014 #8

    Philip Wood

    User Avatar
    Gold Member

    Hello TS. Exactly so. Work done by an external agency per unit charge taking the test charge from infinity to the point is an unnecessary complication; it's very prevalent in the UK. To be fair, this isn't, as I see it, the cause of the OP's confusion. The confusion arose from lack of confidence in using the formal relationship [itex] dV = -E_r dr[/itex], which correctly equates fall in potential to work done per unit charge by the field on the test charge.
     
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